1. Balancing Redox Equations:
Many redox equations can be balanced through trial and error! You have been doing this for a least two years without even knowing it! : )
Do the following:
___Al + ____HCl  ____ AlCl3 + ___ H2
Done!

2. Balancing Redox Reactions:
Another Example: (spectator ions have not been included)
Sn + Ag+  Sn Ag
1. Assign oxidation numbers.

3. 2. Oxidation occurs when the oxidation number increases
2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions.
Oxidation: Sn  Sn2+
Reduction: Ag+  Ag
3. Use electrons to balance the charges in the half reactions. In oxidation the electrons appear on the right. In reduction the electrons appear on the left.

4. Oxidation: Sn  Sn e-
Reduction: Ag+ + 1e-  Ag
4. If the number of electrons transferred is not equal multiple by a whole number so that the number of electrons lost equals the number gained.
Reduction: (Ag+ + 1e- Ag) x2

5. 5. Add the half reactions:
Oxidation: Sn  Sn e-
Reduction: 2Ag+ + 2e-  2Ag
Net Balanced Redox Reaction (notice that the electrons cancel!)
Sn Ag+  Sn Ag

6. Steps for Balancing More:
Step 1: Identify and write the two half-reactions.
Step 2: Balance the elements and charges for each half-reaction:
2 a) Balance all the elements except hydrogen and oxygen.
2b) Balance the oxygen atoms by adding H2O to the appropriate side.
2c) Balance the hydrogen by adding H+ to the appropriate side.
2d) Balance the charges by adding electrons to the appropriate side.
Step 3: Multiply one or both of the half-reactions by a whole number so that the number of electrons gained and lost is equal. Step 4: Combine the half-reactions. Eliminate anything common to the product and reactant sides. Step 5: Check that all of the elements are balanced and that the total charge on each side is the same.

7. Example: Balance: MnO4- + Fe2+  Mn2+ + Fe3+ Step 1:
Reduction: MnO4-  Mn2+ Oxidation: Fe2+  Fe3+
+7,
Step 2:
b) MnO4-  Mn H2O
c) MnO H+  Mn H2O
d) MnO H+ + 5e- Mn H2O

8. Oxidation:
Fe2+  Fe3+ +1e-
Step 3:
MnO H+ + 5e- Mn H2O
(Fe  Fe3+ +1e-) x5

9. Step 4:
Reduction: MnO H+ + 5e- Mn H2O
Oxidation: 5Fe  5 Fe3+ +5e-
Net Reaction: MnO H+ +5Fe2+ Mn H2O Fe3+
(electrons are cancelled!)
Step 5:
Notice that the final redox equation is balanced by atom and by charge.

10. One to try
Redox reactions are usually too complex to use trial and error method.
Balance the following example that is in an acidic solution (assume the presence of H2O and H+):
HNO Fe2+  Fe NO2